Let $c_n$ be the coefficient of $x^{2n}$ in the Maclauren expansion of $\log\cos x$. What can be said about the asymptotics of $c_n$ as $n\to\infty$?
I expect that this question is routine, but I have not studied combinatorics and don't even know where to get started. This is just one small piece of a larger question but I think I can handle the other aspects.
We'll write $[x^n] f(x)$ to mean the coefficient of $x^n$ in the Maclaurin expansion of $f(x)$.
Since
$$ \frac{d}{dx} \log \cos x = - \tan x $$
we have
$$ [x^n] \log \cos x = -\frac{1}{n} [x^{n-1}] \tan x. $$
In particular,
$$ [x^{2n}] \log \cos x = -\frac{1}{2n} [x^{2n-1}] \tan x = - \frac{T_n}{(2n)!}, $$
where $T_n$ is the $n^\text{th}$ tangent number. The asymptotics can then be found by applying the asymptotic formula for the Bernoulli numbers. The resulting asymptotic series is
$$ [x^{2n}] \log \cos x \approx - \frac{2^{2n}-1}{n\pi^{2n}} \left(1+\frac{1}{2^{2n}} + \frac{1}{3^{2n}} + \frac{1}{4^{2n}} + \cdots \right). $$